Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 23, 1998, 181–190

RATIONAL SOLUTIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS

A. Eremenko

Purdue University, Department of Mathematics West Lafayette, Indiana 47907, U.S.A.; eremenko@ math.purdue.edu

Abstract. We prove that degrees of rational solutions of an algebraic differential equation F (dw/dz, w, z) = 0 are bounded. For given F an upper bound for degrees can be determined explicitly. This implies that one can find all rational solutions by solving algebraic equations.

Consider the differential equation

(1)  F (w′, w, z) = 0  (w′ = dw/dz)

where F is a polynomial in three variables.

Theorem 1. For every F there exists a constant C = C(F) such that the degree of every rational solution w of (1) does not exceed C.

This statement is not true for differential equations of higher order. Indeed, all functions w_{n}(z) = z^{n} satisfy

(z \frac{w′}{w} )′ = 0.

We will show that the bound for the degree C(F) can be determined effectively. So theoretically it is possible to find all rational solutions of (1) by substituting an expression for w with indeterminate coefficients and solving the resulting system of algebraic equations. It is a challenging unsolved question whether Theorem 1 can be extended to algebraic solutions. Partial results in this direction were obtained by H. Poincaré in [8], [9].

Before proving the theorem in full generality we give a very simple proof for the particular case when the equation is solved with respect to derivative. This simplified proof does not give any effective bound for C.

1991 Mathematics Subject Classification: Primary 30D35.

Supported by NSF grant DMS 9500636.

Proof in the special case. (The method is similar to [3], see also [4]).

Let us write the equation in the form

F_{1}(w, z)(w′) + F_{0}(w, z) = 0,

where F_{j} are polynomials in w and z for j = 0, 1. Assume that equation (1) has infinitely many rational solutions (otherwise there is nothing to prove). Then we can find three different rational solutions w_{1} , w_{2} and w_{3} such that

(2)  F_{1}(w_{i}(z), z) ≢ 0, i = 1, 2, 3.

Let us consider the finite set E ⊂ C̄ consisting of the following points:

(i) if at some point z_{0} we have F_{1}(w_{i}(z_{0}), z_{0}) = 0 for some i then z_{0} belongs to E;

(ii) the point ∞ and all poles of w_{i}, i = 1, 2, 3 belong to E; (iii) if w_{i}(z_{0}) = w_{j}(z_{0}) for some i ≠ j then z_{0} belongs to E.

Condition (2) guarantees that the subset of E defined in (i) is finite. The subsets of E described in (ii), and (iii) are evidently finite.

Denote by R the set of all rational solutions, different from w_{1} , w_{2} , w_{3} and let w ∈ R. We claim that

(3)  w(z) ≠ w_{i}(z)  for  i = 1, 2, 3  and  z ∈ C̄\E.

Indeed, if for example w(z_{0}) = w_{1}(z_{0}) := w_{0} and z_{0} ∉ E, then w_{0} ≠ ∞ and F_{1}(w_{0}, z_{0}) ≠ 0 in view of (ii) and (i). Thus by the Uniqueness Theorem for solutions of the Cauchy problem we conclude that w = w_{1} , which contradicts to our assumption.

Now we consider the following set of rational functions

S = \{\frac{(w − w_{1})(w_{3} − w_{2})}{(w − w_{2})(w_{3} − w_{1})} : w ∈ R\}.

It follows from (3) and (iii) that functions from S can take the values 0, 1 and ∞ only on E. On the other hand, if f is a rational function of degree d then the preimage f^{−1}({0, 1, ∞}) contains at least d+2 distinct points, which follows from the Riemann–Hurwitz formula. Thus the degrees of functions in S are bounded and so the degrees of functions in R are bounded.

This proof evidently does not provide any algorithm for estimating C(F) for a given polynomial F, or for checking whether a rational solution exists at all. So we give another proof, which permits at least in principle to find the constant C(F) effectively, and which is applicable to all polynomials F. In what follows we will always assume that the polynomial F is irreducible, which does not restrict generality.

Preliminaries and notations. We need some facts from the theory of algebraic functions ([1] or [10, Chapters 18, 19] are standard references) and from differential algebra [7].

Let k = C(z) be the field of rational functions and K be its algebraic closure, that is the field of all algebraic functions. The degree of a rational function has a natural extension to K. Namely for α ∈ K we denote by T (α) the number of poles of α on its Riemann surface (counting multiplicity), divided by the number of sheets of this Riemann surface over C̄. So T is an absolute logarithmic height in the terminology of [6, Chapter III, §1]. Its definition clearly does not depend on the Riemann surface on which α is defined. The following properties are evident:

(4)  T (α^{n}) = nT (α)  for positive integers n  and
(5)  T (α^{−1}) = T (α),  α ∈ K.

For α ∈ K we denote by i(α) the total ramification. That is if a germ of α at z_{0} ∈ C is expressed as

α(z) = ∑ a_{n}(z − z_{0})^{n/m},

(where we assume that m is chosen smallest possible) then this germ contributes m − 1 units to i(α). We have

(6)  T (α′) ≤ 2T (α) + i(α)  ( ′ = d/dz).

Given an irreducible polynomial P ∈ K[t_{1}, t_{2}] we consider the factor R = K[t_{1}, t_{2}]/(P), where (P) is the ideal generated by P. Then R is a field of transcendency degree 1 over K.

A valuation ring V ⊂ R is a ring which contains K, is not identical with R and has the property that for every x ∈ R either x ∈ V or x^{−1} ∈ V . The set of all non-invertible elements of a valuation ring V forms a maximal ideal ν and there is an element t ∈ V , called a local uniformizer, such that ν = tV and ⋂_{n=1}^{∞} t^{n}V = ∅. The factor ring V/ν is equal to K, so every valuation ring defines a map p: R → K ∪ {∞} with the properties:

p(x + y) = p(x) + p(y), p(xy) = p(x)p(y)

whenever the expressions in the right sides of these formulas are defined^{1} and p(α) = α if α ∈ K. A map with such properties is called place. Given a place one can recover the corresponding valuation ring as V_{p} = p^{−1}(K) and there is one-to-one correspondence between places and valuation rings.

1 We use the ordinary conventions: α + ∞ = ∞, α ∈ K and α.∞ = ∞, α ∈ K^{∗} but ∞ + ∞ or 0.∞ are undefined.

Let a place p be given and let V_{p} be its valuation ring. Every element of the field R can be expressed in the form x = t_{p}^{n}u, where u is an invertible element of V_{p}, t_{p} is a local uniformizer and n is an integer. This integer n is called the order of x at the place p and denoted by ord_{p}x.

In the case when R = K(x) the places are in natural one-to-one correspondence with the set K ∪ {∞}, that is p(y) is just the value of the rational function y at the point p ∈ K ∪ {∞}.

Let R_{1} ⊂ R_{2} be a finite field extension where both R_{1} and R_{2} have transcendency degree 1 over K. Then every valuation ring V_{p} ⊂ R_{1} is contained in some valuation ring V_{q} ⊂ R_{2} . We say that this place q in R_{2} lies over the place p in R_{1} . There is at least one but finitely many places in R_{2} lying over a fixed place in R_{1} .

Let t_{1}, t_{2} ∈ R and P be an irreducible polynomial such that P (t_{1}, t_{2}) = 0. If some elements α_{1}, α_{2} ∈ K satisfy P (α_{1}, α_{2}) = 0 then there is a place p in R such that p(t_{1}) = α_{1} and p(t_{2}) = α_{2} .

A divisor is an element of free Abelian group generated by places. If δ = n_{1}p_{1} + ⋯ + n_{q}p_{q} is a divisor then its degree is defined by deg δ = n_{1} + ⋯ + n_{q}. A divisor is called effective if n_{j} ≥ 0 for all j. This defines a partial order relation on the set of divisors: δ_{1} ≥ δ_{2} if δ_{1} −δ_{2} is effective. Every divisor δ can be written as δ = δ^{+} − δ^{−}, where δ^{+} and δ^{−} are effective divisors without common places.

For x ∈ R we denote by δ(x) the divisor ∑(ord_{p}x)p, where summation is spread over all places in R (only finitely many terms in this sum have nonzero coefficients). For every x ∈ R we have deg δ(x) = 0. To every field of transcendency degree 1 corresponds a non-negative integer g, called genus with the following property: for every divisor δ of degree deg δ ≥ g there exists an element x ∈ R such that δ(x) ≥ −δ. This is a corollary from the Riemann–Roch theorem.

Let us recall the construction of the Newton polygon (see, for example [5, IV, §3]). Let x and y be elements of R satisfying an irreducible relation

(7)  P_{0}(x) + P_{1}(x)y + ⋯ + P_{m}(x)y^{m} = 0,  P_{j} ∈ K[x].

Let p_{1}, …, p_{n} be all places in R which lie over some place p in K(x) ⊂ R and t be a local uniformizer at p.

Mark on the plane the points with coordinates (j, ord_{p}P_{j}(x)), 0 ≤ j ≤ m, and consider the maximal convex function whose graph lies below or passes through these points. The slopes of this graph are exactly the numbers −ord_{p_{j}}y/ord_{p_{j}}t.

We use two propositions which follow from consideration of Newton’s polygon.

Proposition 1. The following statements about x and y in (7) are equivalent:

(a) ord_{p}x ≥ 0 implies ord_{p}y ≥ 0 for every place p in R and

(b) deg P_{m} = 0.

Proof. Let p be a place in K(x) and q be a place in R lying over p. Then ord_{p}x ≥ 0 if and only if ord_{q}x ≥ 0. Assume that these inequalities do hold for p and q. We have ord_{p}P_{j}(x) ≥ 0, j = 0, …, m. Then (a) is equivalent to the condition that all slopes of the Newton polygon constructed for p are non-positive. On the other hand, polynomials P_{j} have no common factor because the equation (11) is irreducible, so ord_{p}P_{k}(x) = 0 for some k ∈ {0, …, m}. We conclude that ord_{p}P_{m}(x) = 0 for all places p in K(x) such that ord_{p}x ≥ 0. This is equivalent to (b).

Proposition 2. If δ^{−}(y) ≤ δ^{−}(x) then deg P_{j} ≤ m − j, 0 ≤ j ≤ m.

Proof. Consider the infinite place q in K(x). A local uniformizer at this place is 1/x. We have ord_{q}P_{j}(x) = −deg P_{j}. Our assumption about x and y implies that all slopes of the Newton polygon are at most 1. Furthermore, deg P_{m} = 0 by Proposition 1. We conclude that deg P_{j} ≤ m − j, 0 ≤ j ≤ m.

Lemma 1. If x, y ∈ R and δ^{−}(y) ≤ δ^{−}(x) then there exists a constant C, depending on x and y such that for every place p in R with ord_{p}x ≥ 0 we have

T (p(y)) ≤ T (p(x)) + C.

Proof. Consider the irreducible polynomial relation (7) between x and y. By Proposition 2 we can rewrite (7) in the form

(\frac{y}{x})^{m} + \frac{P_{m−1}(x)}{x}(\frac{y}{x})^{m−1} + ⋯ + \frac{P_{0}(x)}{x^{m}} = 0,  where  deg P_{j} ≤ m − j,

and substitute p(x) and p(y) instead of x and y. It is clear that poles of p(y) can occur only at poles of p(x) or at the poles of coefficients of P_{j}. This proves the lemma.

Lemma 2. Let P ∈ K(t_{1}, t_{2}) be an irreducible polynomial of degree m with respect to t_{1} and of degree n with respect to t_{2} . Given ε > 0 there exists a constant C_{0} depending on P and ε such that for every α and β in K satisfying P (α, β) = 0 we have

(n − ε)T (β) − C_{0} ≤ mT (α) ≤ (n + ε)T (β) + C_{0}.

This is a special case of a general theorem about heights on algebraic varieties [6, Chapter 4, Proposition 3.3]. We give here a simple proof for our special case following the lines of [2].

Proof of Lemma 2. Consider the field R = K[t_{1}, t_{2}]/(P). We have deg δ^{−}(t_{1}) = n and deg δ^{−}(t_{2}) = m. Set s_{1} = t_{1}^{m} and s_{2} = t_{2}^{n}. These elements are connected by an irreducible polynomial relation Q(s_{1}, s_{2}) = 0, which has the same degree with respect to s_{1} and s_{2} . In view of the property (4) it is enough to prove

(8)  T (p(s_{1})) ≤ (1 + ε)T (p(s_{2})) + C

for every place p in R. (The inequality in the opposite direction is then obtained by reversing the roles of s_{1} and s_{2} .)

Choose an integer N so large that

(9)  \frac{N + g}{N} ≤ 1 + ε,

where g is the genus of R.

Consider the divisor δ = (N + g)δ^{−}(s_{2}) − Nδ^{−}(s_{1}) whose degree is equal to gmn ≥ g. By the corollary of the Riemann–Roch theorem mentioned above there is an element x ∈ R such that δ(x) ≥ −δ. It follows that

(N + g)δ^{−}(s_{2}) = δ^{+} ≥ δ^{−}(x),

so by Lemma 1 we conclude that

(10)  (N + g)T (P (s_{2})) ≥ T (p(x)) − C_{1}

for every place p. On the other hand

Nδ^{−}(s_{1}) = δ^{−} ≤ δ^{+}(x) = δ^{−}(x^{−1})

so by Lemma 1 and property (5) we conclude that

NT (p(s_{1})) ≤ T (p(x^{−1})) + C_{2} = T (p(x)) + C_{2}.

Combined with (10) and (9) this gives (8). The lemma is proved.

Now we consider the differential equation (1).

A differential field is a field with an additive map D into itself which satisfies D(xy) = D(x)y + xD(y). Such a map is called derivation. As before we define a field R = K[w′, w]/(F) of transcendency degree 1 over K. There is a unique derivation D: R → R such that D(w) = w′ and D(α) = dα/dz for every α ⊂ K. Here d/dz stands for the usual differentiation in K.

A solution w = α ∈ K of the differential equation F (w′, w) = 0 defines a place p with the additional property that

p(Dx) = \frac{d}{dz} p(x) for every x ∈ R.

We will call such place a differential place.

We say that the differential field R is Fuchsian if the derivation maps every valuation ring into itself. (In [7] such fields are called “differential fields with no movable singularities”.) The following classification of Fuchsian fields can be found in [7]:

1. If R is of genus 0 there is an element x ∈ R such that R = K(x) and Dx = a_{2}x^{2} + a_{1}x + a_{0} with some a_{i} ∈ K.

2. If R is of genus 1, there are two possibilities:

a) There is an element x ∈ R such that R = K(x, Dx) and (Dx)^{2} = a(x − e_{1})(x − e_{2})(x − e_{3}) with some a ∈ K and e_{i} ∈ C, e_{1} + e_{2} + e_{3} = 0 (Poincaré field) or

b) there are x, y ∈ R such that R = K(x, y) and Dx = Dy = 0 (Clairaut field).

3. If R is a Fuchsian field of genus greater then 1 then R is a Clairaut field (see 2a)).

An inspection of the proofs in [7] shows that there is an explicit algorithm of finding the element x, mentioned in 1, 2 or 3. More precisely, if the equation F (w′, w) = 0 is given one can write explicitly the irreducible equation Q(x, w) = 0 where x is the element mentioned in 1, 2 or 3. This also gives explicitly the coefficients a, a_{i}, e_{i}, i = 1, 2, 3 of the differential equations satisfied by x in 1 or 2.

Now we are ready to give the

Proof of Theorem 1. We consider four cases.

Case 1. R is not Fuchsian. This means that there is a place p whose valuation ring V_{p} is not closed with respect to derivation. Let t be a local uniformizer at p. Then ord_{p}Dt < 0. Consider the irreducible equation connecting t and Dt:

(11)  P_{m}(t)(Dt)^{m} + ⋯ + P_{0}(t) = 0,  P_{j} ∈ K[t].

By Proposition 1 we have deg P_{m} > 0. Now put t = x^{−1} + a where a ∈ C is such that P_{j}(a) ≠ 0 for 0 ≤ j ≤ m. From (11) we obtain the equation for x and Dx:

(12)  P_{m}(\frac{1 + ax}{x})x^{−2m}(−Dx)^{m} + ⋯ + P_{0}(\frac{1 + ax}{x}) = 0.

We put d_{j} = deg P_{j} and

(13)  d = max_{j}{d_{j} + 2j} ≥ 2m + 1  (because d_{m} ≥ 1)

After multiplying (1) by x^{d} we obtain an irreducible polynomial equation

(14)  Q_{m}(x)(Dx)^{m} + ⋯ + Q_{0}(x) = 0

where deg Q_{j} = d − 2j. In particular deg Q_{0} = d ≥ 2m + 1 in view of (13).

We have x = r(w, w′) where r is a rational function with coefficients in K. A rational solution α ∈ C(z) of the differential equation F (w′, w) = 0 defines a differential place s and we have s(x) = r(α, dα/dz) so all ramification points of the algebraic function s(x) may come only from the coefficients of r. Thus the total ramification of s(x) is bounded by a constant C_{1} depending only on F and we have by (6)

(15)  T (s(Dx)) ≤ 2T (s(x)) + C_{1}.

On the other hand, applying Lemma 2 to (14) we obtain

(16)  T (s(Dx)) ≥ \frac{2m + 1}{m}T (s(x)) − C_{2}

with some constant C_{2} which also depends only on F. Inequalities (15) and (16) imply that T (s(x)) ≤ m(C_{1} + C_{2}), that is T (s(x)) is bounded by a constant depending only on F. One more application of Lemma 2 shows that the same is true about α = s(w).

Case 2. R is a Fuchsian field of genus 0. In this case it is enough to find a bound for T (α) where α is an algebraic solution of a Riccati differential equation

\frac{dα}{dz} = a_{2}α^{2} + a_{1}α + a_{0}, a_{i} ∈ K.

First we consider the case when a_{2} = 0. Poles of a solution may occur only at the points where a_{1} or a_{0} has a pole. If z_{0} is a pole of α of order n, greater than the order of pole of a_{0} , then a_{1} has at z_{0} a simple pole with residue n. Thus the total number of poles of α, counting multiplicity is bounded from above by a constant depending only on a_{1} and a_{0} .

Now we assume that a_{2} ≠ 0. The substitution

(17)  α = \frac{1}{a_{2}} u − \frac{a_{1}}{2a_{2}} − \frac{1}{2a_{2}^{2}} \frac{da}{dz}

reduces the equation to the standard form form u′ = u^{2} + a with some a ∈ K. Now u may have poles which are not poles of a (they are called movable poles). The residues of u at all movable poles are equal to −1. But the total sum of residues of u(z)dz is equal to 0. Thus to estimate the number of movable poles it is enough to find a bound of residues of u at poles of a. Let z_{0} be a pole of u and a also has a pole at z_{0} . If a has a simple pole at z_{0} then the residue of u at z_{0} is equal to 1. If a has a multiple pole at z_{0} then its multiplicity has to be even, so (assuming that z_{0} is not a ramification point of a and z_{0} ≠ ∞)

a(z) = ∑_{n=−2m}^{∞} b_{n}(z − z_{0})^{n}.

The order of the pole of u at z_{0} should be equal to m, and we substitute the series with indeterminate coefficients

u(z) = ∑_{n=−m}^{∞} c_{n}(z − z_{0})^{n}

into the equation. We obtain:

∑_{n=−m−1}^{∞}(n + 1)c_{n+1}(z − z_{0})^{n} = ( ∑_{n=−m}^{∞} c_{n}(z − z_{0})^{n})^{2} + ∑_{n=−2m}^{∞} b_{n}(z − z_{0})^{n}.

From this equation we see that there are two possible choices for c_{−m} and once c_{−m} is chosen, the coefficients c_{−m+1}, …, c_{−1} are determined in a unique way. Thus the residue c_{−1} has an estimate in terms of a.

If z_{0} is a ramification point of a, let z −z_{0} = ζ^{k}, where ζ is a local parameter on the Riemann surface of a. Then we can rewrite the equation in terms of ζ and v(ζ) = u(z_{0} + ζ^{k}) that is

\frac{dv}{dζ} = kζ^{k−1}(v^{2} + b),

where b(ζ) = a(z_{0} + ζ^{k}), then make again a change of variable similar to (17) to obtain an equation in the standard form and reduce the problem to the case we just considered. The case z_{0} = ∞ is treated similarly with the substitution z = ζ^{−k} so that ζ is a local parameter at ∞. This finishes the proof in Case 2.

Case 3. R is a Poincaré field of genus 1. We have to estimate the number of poles of an algebraic solution u of

(u′)^{2} = a(u − e_{1})(u − e_{2})(u − e_{3}).

The general solution of this equation is given by u = ℘ ∘ A, where ℘ is the Weierstrass elliptic function and A is an Abelian integral

A(z) = \frac{1}{2} ∫ \sqrt{a(z)} dz.

Algebraic solutions u are possible if and only if A is an integral of the first kind (that is \sqrt{a(z)} dz is a holomorphic differential). This implies that

(18)  |a(z)| = O(|z|^{−2−ε}),  for some ε > 0.

Thus, assuming that the Riemann surface S of u has k sheets, we have

∫_{S} \frac{|u′|^{2}}{(1 + |u|^{2})^{2}} dm ≤ ∫_{S} |a|\frac{|(u − e_{1})(u − e_{2})(u − e_{3})|}{(1 + |u|^{2})^{2}} dm ≤ kC,

where C depends on a and e_{i}, and dm stands for the two-dimensional Lebesgue measure pulled back from C to S. The left side of the above formula is the spherical area of the image of S under u, which is equal to π times the total number of poles of u. This implies a bound for T (u) depending only on a.

190 A. Eremenko

Case 4. R is a Clairaut field of genus ≥ 1. Then there is an element x in the field R, transcendental over K with Dx = 0. We have an irreducible polynomial relation Q(w, x) = 0. So for every differential point p with p(w) = α we have Q(α, c) = 0, where c ∈ C, and this gives the desired estimate for T (α) via Lemma 2.

The author thanks A.A. Goldberg and A.Z. Mokhonko for useful discussions, and the referee for his or her valuable comments on the paper.

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Received 1 October 1996